Dielectrics and Capacitance Questions and Answers

Answer: B

The permittivity should be high.

For a parallel-plate capacitor,

\[ C = \frac{\varepsilon_{0}\varepsilon_{r}A}{d} \]

so the capacitance is directly proportional to the permittivity of the dielectric. For a given plate area and spacing, a dielectric with a high relative permittivity, such as a ceramic, gives a much larger capacitance than air.

A low or medium permittivity, or the permittivity of air (\(\varepsilon_{r} \approx 1\)), gives a smaller capacitance, so none of those options achieves a high value.

Answer: A

The field form of Ohm's law is \(\overline{J} = \sigma\overline{E}\).

In electromagnetic theory, Ohm's law is written at each point of a conductor: the current density is proportional to the electric field, with the conductivity \(\sigma\) as the constant of proportionality. Integrating it over a uniform wire of length L and cross-section A gives the circuit form \(V = IR\), with \(R = L/\sigma A\).

V = IR is the circuit law, not the expression for electromagnetic fields the question asks for. \(\sigma = \overline{J}\overline{E}\) multiplies where it should divide, and \(\overline{E} = \overline{J}\sigma\) puts the conductivity where the resistivity \(1/\sigma\) belongs.

Answer: C

The entropy of the universe increases by \(\frac{CV^{2}}{2T}\).

The battery moves charge \(Q = CV\) through emf V, so it does work \(W = QV = CV^{2}\). Half of it, \(\frac{1}{2}CV^{2}\), is stored in the capacitor's field, which is ordered energy carrying no entropy. The other half is dissipated as heat into the surroundings at temperature T:

\[ \begin{aligned} Q_{heat} &= CV^{2} - \frac{1}{2}CV^{2} = \frac{1}{2}CV^{2} \\ \Delta S &= \frac{Q_{heat}}{T} = \frac{CV^{2}}{2T} \end{aligned} \]

\(CV^{2}/T\) counts the whole work as heat, and \(CV^{2}/4T\) and \(CV^{2}/8T\) halve the heat again.

Answer: D

The depolarization field is 0.

The depolarization field is produced by the bound surface charge \(\sigma = \vec{P}\cdot\hat{n}\) on the faces the polarization meets. In a long cylinder polarized along its axis, P is parallel to the curved surface, so \(\vec{P}\cdot\hat{n} = 0\) there; bound charge appears only on the two end faces, which are very far apart and very small compared with the length.

The depolarization factor along the axis of a long needle-like cylinder is therefore zero. The value \(-4\pi P/3\) (Gaussian units) is the depolarization field of a sphere, with factor 1/3, and \(2\pi P\) is the magnitude for a cylinder polarized across its axis, with factor 1/2. A thin slab polarized normal to its face would give \(-4\pi P\).

Answer: A

The induced dipole moment is proportional to E.

A field pulls the electron cloud of a neutral atom one way and the nucleus the other until the restoring force balances the applied force. For ordinary fields the displacement is proportional to the field, so

\[ \vec{p} = \alpha\vec{E} \]

where \(\alpha\) is the atomic polarizability. A simple model of the atom as a point nucleus in a uniform sphere of charge gives \(\alpha = 4\pi\varepsilon_{0}R^{3}\).

Terms in \(E^{2}\) or \(E^{3}\) appear only as small non-linear corrections at very strong fields, and a monoatomic atom has no permanent dipole moment, so the induced moment is not independent of E.

Answer: A

The ratio is \(\epsilon_{1}/\epsilon_{2} = \tan\theta_{1}/\tan\theta_{2}\).

At an uncharged interface the tangential component of E and the normal component of D are continuous:

\[ \begin{aligned} E_{1}\sin\theta_{1} &= E_{2}\sin\theta_{2} \\ \epsilon_{1}E_{1}\cos\theta_{1} &= \epsilon_{2}E_{2}\cos\theta_{2} \end{aligned} \]

Dividing the first equation by the second gives \(\tan\theta_{1}/\epsilon_{1} = \tan\theta_{2}/\epsilon_{2}\), the law of refraction of field lines.

The sine ratio is Snell's law for light, which involves refractive indices rather than permittivities. The cosine and cotangent ratios do not follow from the boundary conditions; the cotangent ratio is in fact the inverse of the correct answer.

Answer: A

The tangential component of the electric field is always continuous at a boundary.

Applying \(\oint \mathbf{E} \cdot d\mathbf{l} = 0\) to a thin rectangular loop straddling the interface gives \(E_{t1} = E_{t2}\), whatever the two media are. The normal component behaves differently: Gauss's law gives \(D_{n1} - D_{n2} = \rho_{s}\), so the normal component of \(\mathbf{E}\) jumps whenever the permittivities differ or a surface charge is present.

'Horizontal' and 'vertical' are not defined relative to the boundary, so they are not the components boundary conditions are written in, and the normal component is continuous only in special cases.